This is Problem 2 from Stein and Shakarchi's Fourier Analysis, Chapter 4. Below is the problem.
Here we present an estimate of Weyl which leads to some interesting results.
(a) Let $S_N = \sum_{n=1}^{N} e^{2\pi if(n)}$. Show that for $H \leq N$, one has $$ |S_N|^2 \leq c \frac{N}{H} \sum_{h=0}^{H} \left|\sum_{n=1}^{N-h} e^{2\pi i\left(f(n+h)-f(n)\right)}\right|,$$
for some constant $c > 0$ independent of $N$, $H$, and $f$.
(b) Use this estimate to show that the sequence $\langle n^2\gamma\rangle$ is equidistributed in $[0, 1)$ whenever $\gamma$ is irrational.
(c) More generally, show that if $\{\xi_n \}$ is a sequence of real numbers so that for all positive integers h the difference $\langle \xi_{n+h} − \xi_n \rangle$ is equidistributed in $[0, 1)$, then $\langle \xi_n \rangle$ is also equidistributed in $[0, 1)$.
(d) Suppose that $P(x) = c_n x^n + \cdots + c_0$ is a polynomial with real coefficients, where at least one of $c_1, \dots, c_n$ is irrational. Then the sequence $\langle P(n)\rangle$ is equidistributed in $[0, 1)$.
Below is my own attempt.
Part (a)
Let $a_n= e^{2\pi if(n)}$ when $1 \leq n \leq N$ and 0 otherwise.
We have, on the one hand, $$H S_N = H \sum_n a_n = \sum_{k=1}^H \sum_n a_{n+k} = \left( (A_1 + A_2 + \cdots + A_H), 1 \right),$$
where $$ \begin{equation} \begin{aligned} A_1 &= (a_1, a_2, \cdots, a_N, 0, 0, 0, \cdots, 0)\\ A_2 &= (0, a_1, a_2, \cdots, a_N, 0, 0, \cdots, 0)\\ A_3 &= (0, 0, a_1, a_2, \cdots, a_N, 0, \cdots, 0)\\ \cdots\\ A_{N+1} &= (0, 0, 0, 0, \cdots, a_1, a_2, \cdots, a_N)\\ 1 &= (1, 1, 1, 1, \cdots, 1, 1, 1, 1,1,1). \end{aligned} \end{equation} $$
These are $2N$-dimensional vectors.
Now we can apply the Cauchy-Schwarz inequality.
$$H^2 |S_N|^2 \leq \big|A_1 + A_2 + \cdots + A_H\big|^2 \cdot |1|^2 = 2N \big|A_1 + A_2 + \cdots + A_H\big|^2 $$
Noticing that $|(A_1,A_{1+j})| = |(A_2,A_{2+j})| = \cdots = |(A_{H-j},A_H)|$ for any $j=0,1,\cdots,H-1$, we have
$$\frac{H^2}{2N} |S_N|^2 \leq \sum_{j=1}^H |A_j|^2 + 2 \sum_{h=1}^{H-1} \sum_{j=1}^{H-h} |(A_j,A_{j+h})| \leq 2H \sum_{h=0}^{H-1} |(A_1,A_{1+h})| $$
On the other hand, we have $$\sum_{h=0}^{H} \left|\sum_{n=1}^{N-h} e^{2\pi i\left(f(n+h)-f(n)\right)}\right| = \sum_{h=0}^{H} |(A_1,A_{1+h})|. $$
Compare the two inequalities above, we have the desired relationship with $c=4$.
Part (b)
Let $f(n) = k n^2 \gamma$ and $S_N$ as defined in (a). We need to show $$\lim_{N\rightarrow\infty} \frac{1}{N} S_N = 0$$ for all $k \neq 0$.
From (a), we have $$ \begin{equation} \begin{aligned} \frac{1}{N^2} |S_N|^2 &\leq c \frac{1}{NH} \sum_{h=0}^{H} \left|\sum_{n=1}^{N-h} e^{2\pi i k (2nh \gamma + h 2\gamma)}\right| \\ &= c \frac{1}{H} \sum_{h=0}^{H} \frac{1}{N}\left|\sum_{n=1}^{N-h} e^{2\pi i k (2nh \gamma)}\right| \\ &= c \frac{1}{H} + c \frac{1}{H} \sum_{h=1}^{H} \frac{1}{N}\left|\sum_{n=1}^{N-h} e^{2\pi i k (2nh \gamma)}\right| \\ \end{aligned} \end{equation} $$
For any $\epsilon > 0$, we can choose an $H$ such that $\frac{c}{H} < \frac{\epsilon}{2}$. Since we already know that $\langle n \cdot 2h\gamma \rangle$ is equidistributed for any non-zero integer $h$, we can choose an $N'$ such that $$\frac{1}{N}\left|\sum_{n=1}^{N-h} e^{2\pi i k (2nh \gamma)}\right| < \frac{1}{H}$$ for all $N>N'$ and all $h \in \{1, 2, \dots, H\}$.
This gives us the desired result.
Part (c)
We can follow exactly the same steps as in part (b).
Part (d)
First assume that $c_1 x$ is the highest degree term which has an irrational coefficient. Define $q$ as the lowest common denominator of $c_n, \cdots, c_2$.
It is easy to show that $$e^{2\pi i k P(n+q)} = e^{2\pi i k \left(P(n)+c_1 q\right)}$$.
We have $$ \begin{equation} \begin{aligned} &\frac{1}{Nq} \sum_{n=1}^{Nq} e^{2\pi i k P(n)} \\ =& \frac{1}{Nq} \sum_{r=1}^q \sum_{s=0}^N e^{2\pi i k P(sq+r)} \\ =& \frac{1}{q} \sum_{r=1}^q e^{2\pi i k P(r)} \frac{1}{N} \sum_{s=0}^N e^{2\pi i k q (s \cdot c_1)}. \end{aligned} \end{equation} $$
This partial sum tends to 0 as $N$ tends to infinity, since $\langle s \cdot c_1 \rangle$ is equidistributed.
From here it is easy to establish $$\lim_{N'\rightarrow\infty}\frac{1}{N'} \sum_{n=1}^{N'} e^{2\pi i k P(n)} = 0$$ for $N'$ which is not necessarily a multiple of $q$.
If the highest degree term of $P(n)$ which has an irrational coefficient has degree $c_m$, the highest degree term of $P(n+h) - P(n)$ has degree at most $c_m-1$. So the rest can be argued by induction and use part (c).